$\sum_{r_n \le x} 2^{-n}$ is continuous on $\Bbb R\setminus \Bbb Q$ and discontinuous on $\Bbb Q$

Question

Let $\{r_n\}_{n\ge 1}$ be an enumeration of the rationals. Define $$F(x) := \sum_{r_n \le x} 2^{-n}$$ Following this answer, I'd like to show that $F$ is discontinuous at every rational, but continuous at every irrational.

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Here's my shot at it:

Pick $x \in \Bbb R\setminus \Bbb Q$ and $\epsilon > 0$. We want $\delta > 0$ such that $|y-x| < \delta$ implies $|F(y) - F(x)| < \epsilon$. Suppose $y \in (x-\delta,x)$. The condition $|F(y) - F(x)| < \epsilon$ is satisfied if and only if $$\left|\sum_{r_n \le y} 2^{-n} - \sum_{r_n \le x} 2^{-n} \right| = \left|\sum_{y < r_n \le x} 2^{-n} \right| < \epsilon$$ I'm not sure what to do next. Could I have any hints, and intuition on why the said function behaves in such a peculiar manner?

Answer 1

Denote $$f_n(x)=\frac1{2^n}\chi_{[r_n, +\infty)}(x),\qquad x\in\mathbb R, n\geq1,$$ then $|f_n|\leq\frac1{2^n}$, and thus $$F(x)=\sum_{n=1}^\infty f_n(x)$$ is a uniformly convergent series. For $x_0\in\mathbb R\setminus\mathbb Q$, $f_n$ is continuous at $x_0$ for each $n$, hence $F$ is continuous at $x_0$. For $x_0\in\mathbb Q$, there exists $n_0$ such that $x_0=r_{n_0}$, then $f_n$ is continuous at $x_0$ for each $n\neq n_0$, hence $$F-f_{n_0}=\sum_{n\neq n_0}f_n$$ is continuous at $x_0$; however, $f_{n_0}$ is discontinuous at $x_0$, hence $F$ is discontinuous at $x_0$.

Answer 2

As $\sum_{n=1}^\infty 2^{-n}$ is a convergent series, for sufficiently large $N$, we must have $\sum_{n=N}^\infty 2^{-n}<\epsilon$.

Let $\delta = \min_{n=1}^N{|x-r_n|}$. If $x$ is irrational, then $\delta>0$, and we have for any $y \in (x-\delta, 0)$, if $y<r_n<x$, then $|r_n-x|<\delta$ hence $n>N$, then $$|F(x)-F(y)|= \sum_{y<r_n\le x} 2^{-n}= \sum_{y<r_n< x} 2^{-n}\le \sum_{n=N}^\infty 2^{-n}<\epsilon$$

Similarly, we can show for any $y\in (x, x+\delta)$, the equality holds as well. Therefore $F(x)$ is continuous at every irrational point.

If $x=r_N$ is rational, for any $y<x$, we must have $$|F(x)-F(y)|= F(x)-F(y)=\sum_{y< r_n \le x } 2^{-n}\ge 2^{-N}$$

no matter how close $y$ is to $x$ from the left.

However, it can be shown that $F(x)$ is right continuous everywhere (More generally this holds for any cumulative probability distribution function).

From the proof (and in particular if you have some experience with probability density), it's easy to see that for any sequence of real numbers $\{r_1, r_2, \cdots\}$, the function $F(x)$ as defined must be continuous everywhere except at every $r_i$. This has little to do with rational or irrational numbers. In particular, $\mathbb Q$ being dense in $\mathbb R$ is irrelevant.